Solving Trig Equations Day 1 HW

• A
• B
• C
• D

Tags

CCSS.HSF.TF.B.7

• No solution
• π/3 +2πn, 5π/3 +2πn
• π/6 +2πn, 11π/6 +2πn
• 2π/3 +2πn, 4π/3 +2πn

Tags

CCSS.HSF.TF.B.7

Solve on the domain [0, 2π)
2sin x cos x = √2 cos x

• π/2, 3π/2
• 0, π/4, 3π/4
• π/2, 3π/2, π/4, 3π/4
• No solution

Tags

CCSS.HSF.TF.B.7

Which of these is equivalent to 2cos²x − 3cosx = 0 ?

• -cos²x = 0

• cosx(2cosx + 3) = 0

• cosx(2cosx − 3) = 0

• cos x = ⅔

Tags

CCSS.HSF.TF.B.7

Which is a correct way to solve the equation

cosx(2cosx − 3) = 0?

• divide cos x from both sides

• set each factor equal to 0 and solve

• distribute cosx into the parentheses

• guess and hope for the best

Tags

CCSS.HSF.TF.B.7

To solve this equation, cos²x + sinx = 1, replace cos²x with

• 1/(sec²x)

• sin²x − 1

• 1 − sin²x

• 1 + tan²x

Tags

CCSS.HSF.TF.C.8

• A
• B
• C
• D

Tags

CCSS.HSF.TF.B.7

• A
• B
• C
• D

Tags

CCSS.HSF.TF.B.7